題目

https://leetcode-cn.com/problems/find-mode-in-binary-search-tree/

題解

中序遍歷二叉搜索樹，可以得到一個有序序列。

遍歷這個有序序列，找到眾數。

import java.util.*;class TreeNode {int val;TreeNode left;TreeNode right;TreeNode() {}TreeNode(int val) {this.val = val;}TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;} }public class Solution {public int[] findMode(TreeNode root) {// 中序遍歷得到有序序列ArrayList<Integer> list = new ArrayList<>();inOrder(root, list);// list 轉 arrayint[] sortedArray = new int[list.size()];for (int i = 0; i < list.size(); i++) {sortedArray[i] = list.get(i);}// 找到眾數（可能包含多個）int base = sortedArray[0];int count = 1;int maxCount = 1;ArrayList<Integer> answer = new ArrayList<>();answer.add(sortedArray[0]);for (int i = 1; i < sortedArray.length; i++) {if (sortedArray[i] == base) {++count;} else {count = 1;base = sortedArray[i];}if (count == maxCount) {answer.add(base);}if (count > maxCount) {maxCount = count;answer.clear();answer.add(base);}}int[] arr = new int[answer.size()];for (int i = 0; i < answer.size(); i++) {arr[i] = answer.get(i);}return arr;}public void inOrder(TreeNode node, List<Integer> list) {if (node == null) return;inOrder(node.left, list);list.add(node.val);inOrder(node.right, list);}// for testpublic static void main(String[] args) {TreeNode t1 = new TreeNode(1);TreeNode t2_1 = new TreeNode(2);TreeNode t2_2 = new TreeNode(2);t1.right = t2_1;t2_1.left = t2_2;int[] arr = new Solution().findMode(t1);for (int i = 0; i < arr.length; i++) {System.out.println(arr[i]);}} }

總結

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